A) \[6\hat{i}+2\hat{j}-3\hat{k}\]
B) \[-18\hat{i}-13\hat{j}+2\hat{k}\]
C) \[18\hat{i}+13\hat{j}-2\hat{k}\]
D) \[6\hat{i}-2\hat{j}+8\hat{k}\]
Correct Answer: B
Solution :
| The relation between linear velocity \[\vec{v},\] angular velocity \[\vec{\omega }\] and position vector \[\vec{r}\] is : |
| \[\vec{v}=\vec{\omega }\times \vec{r}\] |
| \[=(3\hat{j}-4\hat{j}+\hat{k})\times (5\hat{i}-6\hat{j}+6\hat{k})\] |
| \[=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 3 & -4 & 1 \\ 5 & -6 & 6 \\ \end{matrix} \right|\] |
| \[=\hat{i}\left| \begin{matrix} -4 & 1 \\ -6 & 6 \\ \end{matrix} \right|-\hat{j}\left| \begin{matrix} 3 & 1 \\ 5 & 6 \\ \end{matrix} \right|+\hat{k}\left| \begin{matrix} 3 & -4 \\ 5 & -6 \\ \end{matrix} \right|\] |
| \[=(-24+6)\hat{i}-(18-5)\hat{j}+(-18+20)\hat{k}\] |
| \[=-18\hat{i}-13\hat{j}+2\hat{k}\] |
| Alternative: |
| \[\vec{v}=\vec{\omega }\times \vec{r}\] |
| \[=(3\hat{i}-4\hat{j}+\hat{k})\times (5\hat{i}-6\hat{j}+6\hat{k})\] |
| \[=(3\times 5)(\hat{i}\times \hat{i})+[3\times (-6)](\hat{i}\times \hat{j})+(3\times 6)(\hat{i}\times \hat{k})\] |
| \[+(-4\times 5)(\hat{j}\times \hat{i})+(-4\times -6)(\hat{j}\times \hat{j})\] |
| \[+(-4\times 6)(\hat{j}\times \hat{k})+(1\times 5)(\hat{k}\times \hat{i})\] |
| \[+(1\times -6)(\hat{k}\times \hat{j})+(1\times 6)(\hat{k}\times k)\] |
| use \[\hat{i}\times \hat{j}=-\hat{j}\times \hat{i}=\hat{k}\] |
| \[\hat{j}\times \hat{k}=-\hat{k}\times \hat{j}=\hat{i}\] |
| and \[\hat{k}\times \hat{i}=-\hat{i}\times \hat{k}=\hat{j}\] |
| Thus, \[\vec{v}=0+(-18)(\hat{k})+(81)(-\hat{j})\] |
| \[+(-20)(-\hat{k})+0+(-24)(\hat{i})+(5)(\hat{j})+(-6)(-\hat{i})+0\] |
| \[=-18\hat{k}-18\hat{j}+20\hat{k}-24\hat{i}+5\hat{j}+6\hat{i}\] |
| \[=-18\hat{i}-13\hat{j}+2\hat{k}\] |
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